1. Physics II, Pg 1 Physics II Today’s Agenda l Work & Energy. çDiscussion. çDefinition. l Scalar Product. l Work of a constant force. çWork kinetic-energy theorem. l Work of a sum of constant forces. l Work for a sum of displacements with constant force. l Comments. l Look at textbook problems Chp7 - 7,11,12,17,20,23,24,27,29,36

2. Physics II, Pg 2 Work & Energy l One of the most important concepts in physics. çAlternative approach to mechanics. l Many applications beyond mechanics. çThermodynamics (movement of heat). çQuantum mechanics... l Very useful tools. çYou will learn new (sometimes much easier) ways to solve problems. See text: 1-1 and 7-1

3. Physics II, Pg 3 Forms of Energy l Kinetic l Kinetic: Energy of motion. çA car on the highway has kinetic energy. çWe have to remove this energy to stop it. çThe breaks of a car get HOT ! çThis is an example of turning one form of energy into another. (More about this soon)... See text: 1-1 and 7-1

4. Physics II, Pg 4 Forms of Energy l Potential l Potential: Stored, “potentially” ready to use. çGravitational. »Hydro-electric dams etc... çElectromagnetic »Atomic (springs, chemical...) çNuclear »Sun, power stations, bombs... See text: 1-1 and 7-1

5. Physics II, Pg 5 Mass = Energy (but not in Physics II) l Particle Physics: + 5,000,000,000 V e- - 5,000,000,000 V e+ (a) (b) (c) E = 10 10 eV M E = MC 2 ( poof ! )

6. Physics II, Pg 6 Energy Conservation l Energy cannot be destroyed or created. çJust changed from one form to another. energy is conserved l We say energy is conserved ! çTrue for any isolated system. çi.e when we put on the brakes, the kinetic energy of the car is turned into heat using friction in the brakes. The total energy of the “car-breaks-road-atmosphere” system is the same. çThe energy of the car “alone” is not conserved... »It is reduced by the braking. workenergy l Doing “work” on a system will change it’s “energy”... See text: 7-1

7. Physics II, Pg 7 Definition of Work: Ingredients: FS Ingredients: Force ( F ), displacement ( S ) F Work, W, of a constant force F S acting through a displacement S is: F. S W = F. S = FScos(  ) = F S S  F S displacement FSFS “Dot Product” See text: 7-1 and 7-2

8. Physics II, Pg 8 Aside: Scalar Product ( or Dot Product) Definition: ab a. b= abcos(  ) = a[bcos(  )] = ab a = b[acos(  )] = ba b Some properties: abba a. b= b. a abba ba q(a. b) = (qb). a = b. (qa) (q is a scalar) ab cab ac c a. (b + c) = (a. b) + (a. c) (c is a vector) The dot product of perpendicular vectors is 0 !! a abab b baba See text: 7-2

9. Physics II, Pg 9 Aside: Examples of dot products Suppose Then a i j k a = 1 i + 2 j + 3 k b i j k b = 4 i - 5 j + 6 k ab a. b = 1x4 + 2x(-5) + 3x6 = 12 aa a. a = 1x1 + 2x2 + 3x3 = 14 bb b. b = 4x4 + (-5)x(-5) + 6x6 = 77 i i j j k k i. i = j. j = k. k = 1 i j j k k i i. j = j. k = k. i = 0 See text: 7-2

10. Physics II, Pg 10 Aside: Properties of dot products l Magnitude: a 2 = |a| 2 = a. a i j i j = (a x i + a y j ). (a x i + a y j ) i i j j i j = a x 2 ( i. i ) + a y 2 ( j. j ) + 2a x a y ( i. j ) = a x 2 + a y 2 çPythagoras Theorem !! a axax ayay i j See text: 3-5 and 7-2

11. Physics II, Pg 11 Aside: Properties of dot products l Components: a i j ka ia ja k a = a x i + a y j + a z k = (a x, a y, a z ) = (a. i, a. j, a. k ) l Derivatives: çApply to velocity çSo if v is constant (like for UCM): See text: 7-2 since a and v are perpendicular

12. Physics II, Pg 12 Back to the definition of Work: F Work, W, of a force F acting S through a displacement S is: F. S W = F. S F S See text: 7-2

13. Physics II, Pg 13 Work: 1-D Example (constant force) F x A force F = 10N pushes a box across a frictionless floor for a distance  x = 5m. xxxx F byF on FxFx Work done by F on box : W F = F.  x = F  x (since F is parallel to  x) W F = (50 N)x(5m) = 50 N-m. See text: 7-1 and 7-2 See example 7.1

14. Physics II, Pg 14 Units: N-m (Joule) Dyne-cm (erg) = 10 -7 J BTU= 1054 J calorie= 4.184 J foot-lb= 1.356 J eV= 1.6x10 -19 J cgsothermks Force x Distance = Work Newton x [M][L] / [T] 2 Meter = Joule [L] [M][L] 2 / [T] 2 See text: 7-1

15. Physics II, Pg 15 Work & Kinetic Energy: F x A force F = 10N pushes a box across a frictionless floor for a distance  x = 5m. The speed of the box is v 1 before the push, and v 2 after the push. xxxx F v1v1 v2v2i m See text: 7-1 and 7-2

16. Physics II, Pg 16 Work & Kinetic Energy... Fa l Since the force F is constant, acceleration a will be constant. We have shown that for constant a:  v 2 2 - v 1 2 = 2a(x 2 -x 1 ) = 2a  x.  multiply by 1 / 2 m: 1 / 2 mv 2 2 - 1 / 2 mv 1 2 = ma  x  But F = ma 1 / 2 mv 2 2 - 1 / 2 mv 1 2 = F  x xxxx F v1v1 v2v2a i m See text: 7-1, 7-2, 7-5

17. Physics II, Pg 17 Work & Kinetic Energy... l So we find that  1 / 2 mv 2 2 - 1 / 2 mv 1 2 = F  x = W F l Define Kinetic Energy K:K = 1 / 2 mv 2 çK 2 - K 1 = W F (Work kinetic-energy theorem)  W F =  K (Work kinetic-energy theorem) xxxx F v1v1 v2v2a i m

18. Physics II, Pg 18 Work Kinetic-Energy Theorem: NetWork {Net Work done on object} = changekinetic energy {change in kinetic energy of object} l This is true in general: K1K1 K2K2 F F netdS See text: 7-3, 7-4, 7-5

19. Physics II, Pg 19 Work done by Variable Force: (1D) When the force was constant, we wrote W = F  x çarea under F vs x plot: l For variable force, we find the area by integrating: çdW = F(x) dx. F x WgWg xx F(x) x1x1 x2x2 dx See text: 7-3

20. Physics II, Pg 20 A simple application: Work done by gravity on a falling object l What is the speed of an object after falling a distance H, assuming it starts at rest ? FS l W g = F. S = mgScos(0) = mgH W g = mgH Work Kinetic-Energy Theorem: W g = mgH = 1 / 2 mv 2 S gmggmg H j v 0 = 0 v See text: 7-3, 7-4, 7-5

21. Physics II, Pg 21 What about a sum of forces? FFF Suppose F TOT = F 1 + F 2 and the S displacement is S. The work done by each force is: F. S F. S W 1 = F 1. S W 2 = F 2. S W TOT = W 1 + W 2 F. S F. S = F 1. S + F 2. S FF. S = (F 1 + F 2 ). S F. Stotal W TOT = F TOT. S It’s the total force that mattters !! F F TOT S FF1FF1 FF2FF2 See text: 7-3, 7-4, 7-5

22. Physics II, Pg 22 Work by sum of displacements with constant force. W = W 1 + W 2 FSFS = F. S 1 + F. S 2 FSS = F. ( S 1 + S 2 ) FS W = F. S Work depends only on total displacement, not on the “path”. SS1SS1 SS2SS2 S F See text: 7-3, 7-4, 7-5

23. Physics II, Pg 23 Work by sum of displacements with constant force. S F SS1SS1 SS2SS2 SSnSSn SS3SS3 W = W 1 + W 2 +...+ W n FSFSFS = F. S 1 + F. S 2 +... + F. S n FSSS = F. ( S 1 + S 2 +...+ S n ) FS W = F. S Same result as simple case. See text: 7-3, 7-4, 7-5

24. Physics II, Pg 24 Comments: l Time interval not relevant. çRun up the stairs quickly or slowly...same W. FS Since W = F. S l No work is done if: çF çF = 0 or çS çS = 0 or   = 90 o See text: 7-3, 7-4, 7-5

25. Physics II, Pg 25 Comments... FS W = F. S No work done if  = 90 o. T çNo work done by T. çNo work done by N. T v v N See text: 7-3, 7-4, 7-5

26. Physics II, Pg 26 Recap of today’s lecture l Work & Energy. çDiscussion. çDefinition. l Scalar Product. l Work of a constant force. çWork kinetic-energy theorem. l Properties (units, time independence etc). l Work of a combination of forces. l Comments. l Look at textbook problems Serway Chp7 - 7,11,12,17,20,23,24,27,29,36

27. Physics II, Pg 27 Physics II: Lecture Todays Agenda l Review of Work. Last lecture l Work done by gravity near the earths surface. l Examples: çpendulum, inclined plane, free-fall. l Work done by variable force. çSpring l Problem involving spring & friction- Serway 7- 37,39,40,46,51

28. Physics II, Pg 28 Work Kinetic-Energy Theorem: NetWork l {Net Work done on object} l = changekinetic energy l {change in kinetic energy of object} l  W F =  K = 1 / 2 mv 2 2 - 1 / 2 mv 1 2 xxxx F v1v1 v2v2 m W F = F  x See text: 7-3, 7-4, 7-5

29. Physics II, Pg 29 Work done by gravity: FS 1 1 W g = F. S 1 = mgS 1 cos(  1 ) 1 = -mgS 1 cos(  1 ) = -mg  y W g = -mg  y Depends only on  y ! m S1S1S1S1 gmggmg yy 11 11 j See text: 7-1

30. Physics II, Pg 30 Work done by gravity... Fs FS W g =  (F. s) = F. S TOT W g = -mg  y Depends only on  y ! m gmggmg yy j See text: 7-1 See example 7-2 (easy) and example 7-9 (harder)

31. Physics II, Pg 31 Example: Falling Objects v=0 vfvf H vfvf vfvf Free Fall Frictionless incline Pendulum See text: 7-1

32. Physics II, Pg 32 Example: Falling Objects... FS = W = F. S = FScos(  ) No work done if  = 90 o. T çNo work done by T. g çOnly mg does work ! N çNo work done by N. g çOnly mg does work ! gmggmg N v T v mg See text: 7-1 and 7-2

33. Physics II, Pg 33 Example: Falling Objects v=0 vfvf H vfvf vfvf Free Fall Frictionless incline Pendulum Only gravity will do work: W g = mgH = 1 / 2 mv f 2 does not depend on path !! See text: 7-3 and 7-4

34. Physics II, Pg 34 Lifting a book with your hand: What is the total work done on the book ?? l First calculate the work done by gravity: g. S W g = mg. S = -mgS l Now find the work done by the hand: FS W HAND = F HAND. S = F HAND S gmggmg S F F HAND v v = const a a = 0 See text: 7-1 to 7-5

35. Physics II, Pg 35 Example: Lifting a book... W g = -mgS W HAND = F HAND S W TOT = W HAND + W g = F HAND S -mgS = (F HAND -mg)S a = 0 since a = 0 l So W TOT = 0 !! gmggmg S F F HAND v v = const a a = 0 See text: 7-1 to 7-5

36. Physics II, Pg 36 Example: Lifting a book... Work Kinetic-Energy Theorem says: W F =  K NetWorkchangekinetic energy {Net Work done on object}={change in kinetic energy of object} v In this case, v is constant so  K = 0 and so W F must be 0, as we found. gmggmg S F F HAND v v = const a a = 0 See text: 7-5

37. Physics II, Pg 37 Work done by Variable Force: (1D) When the force was constant, we wrote W = F  x çarea under F vs x plot: l For variable force, we find the area by integrating: çdW = F(x) dx. F x WgWg xx F(x) x1x1 x2x2 dx See text: 7-3

38. Physics II, Pg 38 1-D Variable Force Example: Spring l For a spring we know that F x = -kx. F(x) x2x2 x x1x1 -kx equilibrium F = - k x 1 F = - k x 2 See text: 7-3

39. Physics II, Pg 39 Spring... l The work done by the spring W s during a displacement from x 1 to x 2 is the area under the F(x) vs x plot between x 1 and x 2. WsWs F(x) x2x2 x x1x1 -kx equilibrium See text: 7-3

40. Physics II, Pg 40 Spring... F(x) x2x2 WsWs x x1x1 -kx l The work done by the spring W s during a displacement from x 1 to x 2 is the area under the F(x) vs x plot between x 1 and x 2. See text: 7-3

41. Physics II, Pg 41 Problem: Spring pulls on mass. l A spring (constant k) is stretched a distance d, and a mass m is hooked to its end. The mass is released (from rest). What is the speed of the mass when it returns to the equilibrium position if it slides without friction? equilibrium position stretched position (at rest) d after release back at equilibrium position vEvE v m m m m See text: 7-3

42. Physics II, Pg 42 Problem: Spring pulls on mass. l First find the net work done on the mass during the motion from x=d to x=0 (only due to the spring): stretched position (at rest) d equilibrium position veve m m i See text: 7-3

43. Physics II, Pg 43 Problem: Spring pulls on mass. l Now find the change in kinetic energy of the mass: stretched position (at rest) d equilibrium position veve m m i See text: 7-3

44. Physics II, Pg 44 Problem: Spring pulls on mass. Now use work kinetic-energy theorem: W net = W S =  K. stretched position (at rest) d equilibrium position veve m m i See text: 7-3

45. Physics II, Pg 45 Problem: Spring pulls on mass. Now suppose there is a coefficient of friction  between the block and the floor? fS The total work done on the block is now the sum of the work done by the spring W S (same as before) and the work done by friction W f. W f = f. S = -  mg d stretched position (at rest) d equilibrium position veve m m i f =  mg S See text: 7-3

46. Physics II, Pg 46 Problem: Spring pulls on mass. Again use W net = W S + W f =  K W f = -  mg d stretched position (at rest) d equilibrium position veve m m i f =  mg S See text: 7-3

47. Physics II, Pg 47 Recap of today’s lecture l Review l Work done by gravity near the earths surface. l Examples: çpendulum, inclined plane, free-fall. l Work done by variable force. çSpring l Problems involving spring & friction. Serway 7- 37,39,40,46,51

48. Physics II, Pg 48

49. Physics II, Pg 49 Physics II: Lecture Todays Agenda l Review. l Work done by variable force in 3-D. çNewtons gravitational force. l Conservative Forces & Potential energy. l Conservation of “Total Mechanical Energy” çExample: Pendulum. l Nonconservative force çfriction l General work-energy theorem l Example problem. l Serway Chp 8- 1,2,7,9,11,12,14,20,22,25,27,32,35

50. Physics II, Pg 50 Work by variable force in 3-D: F l Work dW F of a force F acting through an infinitesmal dS displacement dS is: FdS dW = F. dS l The work of a big displacement through a variable force will be the integral of a set of infinitesmal displacements: FdS W TOT = F. dSFdS See text: 7-4

51. Physics II, Pg 51 Work by variable force in 3-D: Newtons Gravitational Force dS l Work dW g done on an object by gravity in a displacement dS is given by: Fds rr r dW g = F g. ds = (-GMm / R 2 r ). (dR r + Rd  ) dW g = (-GMm / R 2 ) dR ( since r.  = 0 ) ^ ^ ^ r  ^ ^ ds Rd  dR R FFgFFg m M dd ^ ^ See text: 7-4, 8-1 Looking at sections 9-2 and 9-4 in the text might help.

52. Physics II, Pg 52 Work by variable force in 3-D: Newtons Gravitational Force l Integrate dW g to find the total work done by gravity in a “big” displacement: W g = dW g = (-GMm / R 2 ) dR = GMm (1/R 2 - 1/R 1 ) F F g (R 1 ) R1R1 R2R2 F F g (R 2 ) R1R1 R2R2 R1R1 R2R2 m M Looking at sections 9-2 and 9-4 in the text might help. See text: 7-4, 8-1

53. Physics II, Pg 53 Work by variable force in 3-D: Newtons Gravitational Force not on path taken l Work done depends only on R 1 and R 2, not on path taken. R1R1 R2R2 m M See text: 7-4, 8-1 Looking at sections 9-2 and 9-4 in the text might help.

54. Physics II, Pg 54 Newtons Gravitational Force Near the Earths Surface: Suppose R 1 = R E and R 2 = R E +  y but we have learned that So: W g = -mg  y R E +  y M m RERE See text: 7-4 See example 7-9

55. Physics II, Pg 55 Conservative Forces: l We have seen that the work done by gravity does not depend on the path taken, only on the change in separation between the two objects. R1R1 R2R2 M m h m W g = -mgh See text: 8-1 See example 7-9

56. Physics II, Pg 56 Conservative Forces: conservative l In general, if the work done does not depend on the path taken, the force involved is said to be conservative. l Gravity is a conservative force : l A spring produces a conservative force: l Any constant force is also conservative: S F See text: 8-1 and 8-2

57. Physics II, Pg 57 Potential Energy l For any conservative force F we can define a potential energy function U in the following way: çThe work done by a conservative force equals (-) the change in the potential energy function. l This can be written as: FS W = F. dS = -  U FS  U = U 2 - U 1 = - W = - F. dS S1S1 S2S2 S1S1 S2S2 U2U2 U1U1 See text: 8-2

58. Physics II, Pg 58 Gravitational Potential Energy We have seen that the work done by gravity near the earths surface when an object of mass m is lifted a distance  y is W g = -mg  y. The change if potential energy of this object is therefore:  U = - W g = mg  y. yy m W g = -mg  y j See text 8-2 See example 8-1

59. Physics II, Pg 59 Gravitational Potential Energy So we see that the change in U near the earths surface is:  U = - W g = mg  y = mg(y 2 -y 1 ). arbitrary constant l So U = mgy + U 0 where U 0 is an arbitrary constant. l Having an arbitrary constant U 0 is equivalent to saying that we can choose the y location where U = 0 to be anywhere we want to. y1y1 m W g = -mg  y j y2y2 See text: 8-2 See example 8-1

60. Physics II, Pg 60 Potential Energy Recap: l For any conservative force we can define a potential energy function U such that: l The potential energy function U is always defined only up to an additive constant. çYou can choose the location where U = 0 to be anywhere convenient. FS  U = U 2 - U 1 = - W = - F. dS S1S1 S2S2 See text: 8-1 and 8-2

61. Physics II, Pg 61 Conservative Forces & Potential Energies (stuff you should know): ForceF Work W(1-2) Change in P.E  U = U 2 - U 1 P.E. function U Fj F g = -mg j Fr F g = r F s = -kx ^ ^ -mg(y 2 -y 1 ) mg(y 2 -y 1 ) mgy + C

62. Physics II, Pg 62 Conservation of Energy l If only conservative forces are present, the total energy (sum of potential and kinetic energies) of a system is conserved (i.e. constant). constant!!! l If only conservative forces are present, the total energy (sum of potential and kinetic energies) of a system is conserved (i.e. constant). E = K + U is constant !!! l Both K and U can change as long as E = K + U is constant. See text: 8-3

63. Physics II, Pg 63 Example: The simple pendulum. l Suppose we release a bob or mass m from rest a distance h 1 above it’s lowest possible point. çWhat is the maximum speed of the bob and where does this happen ? çTo what height h 2 does it rise on the other side ? v h1h1 h2h2 m See text: 8-3 See example A Pendulum

64. Physics II, Pg 64 Example: The simple pendulum. l Energy is conserved since gravity is a conservative force (E = K + U is constant) l Choose y = 0 at the original position of the bob, and U = 0 at y = 0 (arbitrary choice). E = 1 / 2 mv 2 + mgy. v h1h1 h2h2 y y=0 See example, A Pendulum See text: 8-3

65. Physics II, Pg 65 Example: The simple pendulum. l E = 1 / 2 mv 2 + mgy. çInitially, y = 0 and v = 0, so E = 0. çSince E = 0 initially, E = 0 always since energy is conserved. y y=0 See text: 8-3 See example, A Pendulum

66. Physics II, Pg 66 Example: The simple pendulum. l E = 1 / 2 mv 2 + mgy. l So at y = -h, E = 1 / 2 mv 2 - mgh = 0. 1 / 2 mv 2 = mgh l 1 / 2 mv 2 will be maximum when mgh is minimum. l 1 / 2 mv 2 will be maximum at the bottom of the swing ! y y=0 y=-h h See text: 8-3 See example, A Pendulum

67. Physics II, Pg 67 Example: The simple pendulum. l 1 / 2 mv 2 will be maximum at the bottom of the swing ! l So at y = -h 1 1 / 2 mv 2 = mgh 1 v 2 = 2gh 1 v h1h1 y y=0 y=-h 1 See text: 8-3 See example, A Pendulum

68. Physics II, Pg 68 Example: The simple pendulum. l Since 1 / 2 mv 2 - mgh = 0 it is clear that the maximum height on the other side will be at y = 0 and v = 0. l The ball returns to it’s original height. y y=0 See text: 8-3 See example, A Pendulum

69. Physics II, Pg 69 Example: The simple pendulum. l The ball will oscillate back and forth. The limits on it’s height and speed are a consequence of the sharing of energy between K and U. E = 1 / 2 mv 2 + mgy = K + U = 0. y See text: 8-3 See example A Pendulum

70. Physics II, Pg 70 Example: Airtrack & Glider l A glider of mass M is initially at rest on a horizontal frictionless track. A mass m is attached to it with a massless string hung over a massless pulley as shown. What is the speed v of M after m has fallen a distance d ? d M m v v See text: 8-1 to 8-3

71. Physics II, Pg 71 Example: Airtrack & Glider l Energy is conserved l Energy is conserved since all forces are conservative. Choose initial configuration to have U=0.  K = -  U d M m v lets check this See text: 8-1 to 8-3

72. Physics II, Pg 72 Problem: Hotwheel. l A toy car slides on the frictionless track shown below. It starts at rest, drops a distance d, moves horizontally at speed v 1, rises a distance h, and ends up moving horizontally with speed v 2. çFind v 1 and v 2. h d v1v1 v2v2 See text: 8-4

73. Physics II, Pg 73 Problem: Hotwheel... Energy is conserved, so  E = 0  K = -  U Moving down a distance d,  U = -mgd,  K = 1 / 2 mv 1 2 l Solving for the speed: h d v1v1 See text: 8-4

74. Physics II, Pg 74 Problem: Hotwheel... l At the end, we are a distance d-h below our starting point.  U = -mg(d-h),  K = 1 / 2 mv 2 2 l Solving for the speed: h d v2v2 d-h See text: 8-4

75. Physics II, Pg 75 Non-conservative Forces: l If the work done does not depend on the path taken, the force involved is said to be conservative. l If the work done does depend on the path taken, the force involved is said to be non-conservative. l An example of a non-conservative force is friction: l Pushing a box across the floor, the amount of work that is done by friction depends on the path taken. çWork done is proportional to the length of the path ! See text: 8-6

76. Physics II, Pg 76 Non-conservative Forces: Friction Suppose you are pushing a box across a flat floor. The mass of the box is m and the kinetic coefficient of friction is . F. D The work done in pushing it a distance D is given by: W f = F f. D = -  mgD. D F f = -  mg See text: 8-6 See example 8-9

77. Physics II, Pg 77 Non-conservative Forces: Friction Since the force is constant in magnitude, and opposite in direction to the displacement, the work done in pushing the box through an arbitrary path of length L is just W f = -  mgL. l Clearly, the work done depends on the path taken. l W path 2 > W path 1. A B path 1 path 2 See text: 8-6

78. Physics II, Pg 78 Generalized Work Energy Theorem: l Suppose F NET = F C + F NC (sum of conservative and non- conservative forces). l The total work done is: W TOT = W C + W NC The Work Kinetic-Energy theorem says that: W TOT =  K.  W TOT = W C + W NC =  K But W C = -  U So W NC =  K +  U =  E

79. Physics II, Pg 79 Generalized Work Energy Theorem: l The change in total energy of a system is equal to the work done on it by non-conservative forces. E of system not conserved !  If all the forces are conservative, we know that energy is conserved:  K +  U =  E = 0 which says that W NC = 0, which makes sense. çIf some non-conservative force (like friction) does work, energy will not be conserved by an amount equal to this work, which also makes sense. W NC =  K +  U =  E

80. Physics II, Pg 80 Problem: Block Sliding with Friction A block slides down a frictionless ramp. Suppose the horizontal (bottom) portion of the track is rough, such that the kinetic coefficient of friction between the block and the track is . çHow far, x, does the block go along the bottom portion of the track before stopping ? x d  See text: 8-6

81. Physics II, Pg 81 Problem: Block Sliding with Friction... Using W NC =  K +  U As before,  U = -mgd W NC = work done by friction = -  mgx.  K = 0 since the block starts out and ends up at rest. W NC =  U-  mgx = -mgd x = d /  x d  See text: 8-6

82. Physics II, Pg 82 Recap of today’s lecture l Work done by variable force in 3-D. (Text: 7-4) çNewtons gravitational force. (Text: 8-1, 9-2, 9-4) l Conservative Forces & Potential energy. (Text: 8-1 to 8-5) l Conservation of “Total Mechanical Energy” çExamples: Pendulum, Airtrack, Hotwheel. l Nonconservative force (Text: 8-1 to 8-6) çfriction l General work-energy theorem l Example problem. l Look at Textbook problems Chapter 8: 1,2,5,9,10,11,14,15,17

83. Physics II, Pg 83 Physics II Todays Agenda l Problems using work-energy theorem. çSpring shot. çEscape velocity. çLoop the loop. çVertical springs. l Definition of Power, with example

84. Physics II, Pg 84 Problem: Spring Shot l A sling shot is made from a pair of springs each having spring constant k. The initial length of each spring is x 0. A puck of mass m is placed at the point connecting the two springs and pulled back so that the length of each spring is x 1. The puck is released. What is it’s speed v after leaving the springs? (The relaxed length of each spring is x R ). v x0x0 x1x1 mmm xRxR See text: 7-3

85. Physics II, Pg 85 Problem: Spring Shot Only conservative forces are at work, so energy is conserved. E I = E F  K = -  U s x0x0 x1x1 mm See text: 7-3

86. Physics II, Pg 86 Problem: Spring Shot Only conservative forces are at work, so energy is conserved. E I = E F  K = -  U s v mmat rest See text: 7-3

87. Physics II, Pg 87 Problem: Spring Shot Only conservative forces are at work, so energy is conserved. E I = E F  K = -  U s v m m See text: 7-3

88. Physics II, Pg 88 Problem: How High? l A projectile of mass m is launched straight up from the surface of the earth with initial speed v 0. What is the maximum distance from the center of the earth R MAX it reaches before falling back down. R MAX RERE v0v0 m M See text: 9-1 to 9-4

89. Physics II, Pg 89 Problem: How High... l No non-conservative forces: çW NC = 0   K = -  U R MAX v0v0 m h MAX l And we know: RERE M See text: 9-4

90. Physics II, Pg 90 R MAX RERE v0v0 m h MAX M Problem: How High... See text: 9-4

91. Physics II, Pg 91 Escape Velocity l If we want the projectile to escape to infinity we need to make the denominator in the above equation zero: We call this value of v 0 the escape velocity v esc See text: 9-4

92. Physics II, Pg 92 Escape Velocity l Remembering that we find the escape velocity from a planet of mass M p and radius R p to be: (where G = 6.67 x 10 -11 m 3 kg -1 s -2 ). Moon Earth Sun Jupiter R p (m)M p (kg) g p (m/s 2 )v esc (m/s) 6.378(10) 6 5.977(10) 24 1.738(10) 6 7.352(10) 22 7.150(10) 7 1.900(10) 27 6.960(10) 8 1.989(10) 29 9.80 1.62 24.8 27.4 11.2(10) 3 2.38(10) 3 59.5(10) 3 195.(10) 3

93. Physics II, Pg 93 Problem: Space Spring l A low budget space program decides to launch a 10,000 kg space-ship into space using a big spring. If the space ship is to reach a height R E above the surface of the earth, what distance d must the launching spring be compressed if it has a spring constant of 10 8 N/m. d k See text: 7-3 and 9-4

94. Physics II, Pg 94 Problem: Space Spring... l Since gravity is a conservative force, energy is conserved. Since K = 0 both initially and at the maximum height (v=0) we know: l U before = U after l { U S + U G } before = { U G } after See text: 7-3 and 9-4

95. Physics II, Pg 95 Problem: Space Spring l For the numbers given, d = 35.3 m l But don’t get too happy... çF = kd = ma ça = kd/m ça = 1.8 x 10 6 m/s 2. ça = 180000 g çastronaut unhappy d k a So we find See text: 7-3 and 9-4

96. Physics II, Pg 96 Problem: Loop the loop l A mass m starts at rest on a frictionless track a distance H above the floor. It slides down to the level of the floor where it encounters a loop of radius R. What is H if the mass just barely makes it around the loop without losing contact with the track. H R

97. Physics II, Pg 97 Problem: Loop the loop l Draw a FBD of the mass at the top of the loop: l F j l F TOT = -(mg+N) j a j l ma = -mv 2 /R j l If it “just” makes it, N = 0. çmg = mv 2 /R H R v mg N v i j

98. Physics II, Pg 98 Problem: Loop the loop Now notice that energy is conserved.  K = -  U.  U = -mg(h) = -mg(H-2R),  K = 1 / 2 mv 2 = 1 / 2 mRg l mg(H-2R) = 1 / 2 mRg H R h = H - 2R v

99. Physics II, Pg 99 Vertical Springs l A spring is hung vertically, it’s relaxed position at y=0 (a). When a mass m is hung from it’s end, the new equilibrium position is y E (b). l Free Body Diagram of mass in case (b) lets us relate k and y E : -ky E - mg = 0(y E < 0) mg = -ky E y = 0 y = y E j k m (a) (b) (ok since y E is a negative number) See text: 7-3 Recall example 7-7, Bungee Jumping

100. Physics II, Pg 100 Vertical Springs l The potential energy of the spring- mass system is: y = 0 y = y E j k m (a) (b) but mg = -ky E choose C to make U=0 at y = y E : Recall example 7-7, Bungee Jumping See text: 7-3, 8-2

101. Physics II, Pg 101 Vertical Springs l So: y = 0 y = y E j k m (a) (b) which can be written: See text: 7-3, 8-2

102. Physics II, Pg 102 Vertical Springs l So if we define a new y’ co-ordinate system such that y’ = 0 is at the equilibrium position, ( y’ = y - y E ) then we get the simple result: y’ = 0 j k m (a) (b) See text: 7-3, 8-2

103. Physics II, Pg 103 Vertical Springs l If we choose y = 0 to be at the equilibrium position of the mass hanging on the spring, we can define the potential in the simple form. l Notice that g does not appear in this expression !! çBy choosing our coordinates and constants cleverly, we can hide the effects of gravity. y = 0 j k m (a) (b) See text: 7-3, 8-2

104. Physics II, Pg 104 U of Spring -60 -40 -20 0 20 40 60 80 100 120 140 160 -10-8-6-4-20246810 y U U S = 1 / 2 ky 2 See text: 8-2 See Figure 8-9

105. Physics II, Pg 105 U of Gravity -60 -40 -20 0 20 40 60 80 100 120 140 160 -10-8-6-4-20246810 y U U G = mgy See text: 8-2 See Figure 8-7

106. Physics II, Pg 106 U of Spring + Gravity -60 -40 -20 0 20 40 60 80 100 120 140 160 -10-8-6-4-20246810 y U U G = mgy U S = 1 / 2 ky 2 U TOT =U G + U S shift due to mgy term See text: 8-2

107. Physics II, Pg 107 U of Spring + Gravity -60 -40 -20 0 20 40 60 80 100 120 140 160 -10-8-6-4-20246810 y U U S = 1 / 2 ky 2 U TOT =U G + U S + C shift due to mgy term See text: 8-2

108. Physics II, Pg 108 Vertical Springs: Example Problem l If we displace the mass a distance d from equilibrium and let it go, it will oscillate up & down. Relate the maximum speed of the mass v M to d and the spring constant k. l Since all forces are conservative, E = K + U is constant. çWe know: y = 0 j k m y = d y = -d vMvM See text: 7-3, 8-2

109. Physics II, Pg 109 Vertical Springs: Example Problem l At the initial stretched position and K = 0 (since v=0). l Since E is conserved, will always be true ! l Energy is shared between the K and U terms. çAt y = d or -d the energy is all potential çAt y = 0, the energy is all kinetic. y = 0 j k m y = d y = -d vMvM See text: 7-3, 8-2

110. Physics II, Pg 110 Power FS l We have seen that W = F. dS çThis does not depend on time ! l Power is the “rate of doing work”: FSFv Fv l If the force does not depend on time: dW/dt = F. dS/dt = F. v P = F. v l Units of power: J/sec = Nm/sec = WattsFS v See text: 7-6

111. Physics II, Pg 111 Power l A 2000kg trolley is pulled up a 30 degree hill at 20 mi/hr by a winch at the top of the hill. How much power must the winch be producing ? Fv Tv l The power is P = F. v = T. v l Since the trolley is not accelerating, the net force on it must be zero. In the x direction:  T - mg sin(  ) = 0  T = mg sin(  )  v mg T winch x y See text: 7-6

112. Physics II, Pg 112 Power Tv Tv l P = T. v = Tv since T is parallel to v So P = mgv sin(  ) v = 20 mi/hr = 8.93 m/s g = 9.8 m/s 2 m = 2000 kg sin(  ) = sin(30 o ) = 0.5 and P = (2000 kg)(9.8 m/s 2 )(8.93 m/s)(0.5) = 88,000. W  v mg T winch x y See text: 7-6

113. Physics II, Pg 113 Recap of today’s lecture l Problems using work-energy theorem. çSpring shot. çEscape velocity. çLoop the loop. çVertical springs. l Definition of Power, with example l Look at textbook problems